Quantum microcanonical ensemble: Entrophy

Question

I'm studying quantum microcanonical ensemble, in particular an ideal system of $N$ particles with angular momentum $j=1/2$ which interact with an external magnetic field $B$. The entrophy $S$ of the system is

$$S=Nk_B\left[\left(\frac{E}{2N\mu B}-\frac{1}{2}\right)\ln\left(\frac{1}{2}-\frac{E}{2N\mu B}\right)-\left(\frac{E}{2N\mu B}+\frac{1}{2}\right)\ln\left(\frac{1}{2}+\frac{E}{2N\mu B}\right)\right].$$

It was obtained by

$$S = k_B \ln \omega(E;N;B)$$

Then, I studied that this expression of the entrophy has a maximum for $E=0$ and the minimums are locatted in $E=-\mu B N$ and $E=\mu B N$. I also found the equation for the temperature

$$T=\frac{2\mu B}{k_B}\frac{1}{\ln\left(\frac{NB\mu-E}{NB\mu +E}\right)},$$

which is unbounded in $\mathbb{R}$ like you can see easily.

However, I don't know if I the definition of the entrophy

$$S = k_B \ln \Phi(E;N;B)$$

is equivalent to the other.

I look for the answer to this question and I find it in a book. It says that they are different because in $E\geq 0$ the number of quantum states don't grow when the energy grows up. Despite I have this answer, I don't understand it very much.

What does it mean with it?

Notation:

$\omega(E,N,B)$ denote the number of leveles with energy $E-\frac{\Delta E}{2}<E_n<E+\frac{\Delta E}{2}$

$\Phi(E,N,B)$ denote $\int \omega(E,N,B)dE$

Answer 1

If I understand your question correctly, they are equivalent. Proof:

Say $\Omega$ is the number of microstates up to energy $E$ (from $E=0$ to $E = E'$). Then $\omega=\frac{d\Omega}{dE}$ is a density of states.

The number of microstates with energy $E'<E<E'+\delta E$ should be $W = \omega \delta E$, you can think of it as a shell in phase space.

You know entropy depends on the logarithm of $W$. Its

$$ S = k\log W $$

you want to show if thats equivalent to

$$ k\log \Omega $$

so what you need to see is that the difference in the two methods of calculation is very small.

First the number of microstates up to energy $E'$ which is $\Omega(E')$ is less than $\omega(E') \times E'$. You can see this because $\omega$ is always an increasing function of $E$ which implies that $\Omega = \int \omega dE$ is necessarily less than $\Omega\times E'$ otherwise there would've been a point where $\omega$ had a negative slope, which cant be.

The difference you need to calculate is then

$$ S_1 - S_2 = k (log\omega E'-\log \omega \delta E) = k \log \frac{E'}{\delta E} $$

if you consider that entropy is a quantity proportional to $N$ which is of order $10^{23}$ then clearly this quantity is negligible.

So you can see that both quantities should give the same answer, BUT you have to exclude certain negligible terms.

Note that I imply different meanings to the same notation you use (the symbol $\omega$ ). I've explained what they mean.

The conclusion is that you can calculate entropy using the number of microstates, the derivative of the number of microstates up to some energy, or the number of microstates up to some energy, considering they give very small differences.